Re: [math-fun] Maximum number of triples of orthogonal directions

Just using floor( http://oeis.org/A006855 / 3 ) gives an upper bound on f(N).

From it we find

f(3)=1, f(4)=1, f(5)=2, f(6)=2, f(7)=3, f(8)=3, f(9)=4, f(10)=5. The first open case now is f(11)=5 or 6. [If the graph G arising from the Petersen graph P by using vertices(G)=EdgeMidpoints(P) and draw a G-triangle for each P-vertex, is realizable, that would prove f(15)=10.] -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)